a) \(f\left( 1 \right) = 3.1 = 3;f\left( { - 2} \right) = 3.\left( { - 2} \right) =  - 6;f\left( {\dfrac{1}{3}} \right) = 3.\dfrac{1}{3} = 1\).

b) Ta có: \(f\left( { - 3} \right) = 3.\left( { - 3} \right) =  - 9;f\left( { - 1} \right) = 3.\left( { - 1} \right) =  - 3\)

\(f\left( 0 \right) = 3.0 = 0;f\left( 2 \right) = 3.2 = 6;f\left( 3 \right) = 3.3 = 9\);

Ta lập được bảng sau

Cho hàm số y=f(x)y=f(x) có đạo hàm và liên tục trên [0;π2][0;π2]thoả mãn f(x)=f′(x)−2cosxf(x)=f′(x)−2cosx. Biết f(π2)=1f(π2)=1, tính giá trị f(π3)f(π3)

A. √3+1/2         B. √3−1/2          C. 1−√3/2             D. 0

B

\(f'\left(x\right)-f\left(x\right)=2cosx\)

\(\Leftrightarrow e^{-x}.f'\left(x\right)-e^{-x}.f\left(x\right)=2e^{-x}cosx\)

\(\Rightarrow\left[e^{-x}.f\left(x\right)\right]'=2e^{-x}.cosx\)

Lấy nguyên hàm 2 vế:

\(\Rightarrow e^{-x}.f\left(x\right)=\int2e^{-x}cosxdx=e^{-x}\left(sinx-cosx\right)+C\)

Thay \(x=\dfrac{\pi}{2}\Rightarrow e^{-\dfrac{\pi}{2}}.1=e^{-\dfrac{\pi}{2}}+C\Rightarrow C=0\)

\(\Rightarrow f\left(x\right)=sinx-cosx\)

\(\Rightarrow f\left(\dfrac{\pi}{3}\right)=\dfrac{\sqrt{3}-1}{2}\)

Ta có y = f(x) = 3x² + 1. Do đó

f(\(\dfrac{1}{2}\)) = 3.\(\left(\dfrac{1}{2}\right)^2\) + 1 = \(\dfrac{3}{4}\)+ 1 = \(\dfrac{7}{4}\)

f(1) = 3.1² + 1 = 3.1 + 1 = 3 + 1 = 4

f(3) = 3.3² + 1 = 3.9 + 1 = 27 + 1 = 28.

$y=f\left(x\right)=3x^2+1$

$\mathrm{f\; \backslash (\backslash left(\backslash dfrac\{1\}\{2\}\backslash right)\backslash )=\; 3\; .\; \backslash (\backslash left(\backslash dfrac\{1\}\{2\}\backslash right)^2\backslash )\; +\; 1\; =\; 3\; .\; \backslash (\backslash dfrac\{1\}\{4\}\backslash )\; +\; 1\; =\; \backslash (\backslash dfrac\{3\}\{4\}+1\backslash )\; =\; \backslash (\backslash dfrac\{7\}\{4\}\backslash )}$

f (1) = 3 . 1² + 1= 3 + 1 = 4

f (3) = 3 . 3² + 1 = 3 . 9 + 1 = 28

Ta có : y = f(x) = 3\(x^2+1\)

Khi đó : f\(\left(\dfrac{1}{2}\right)=3.\left(\dfrac{1}{2}\right)^2+1=\dfrac{3}{4}+1=\dfrac{7}{4}\)

\(f\left(1\right)=3.\left(1\right)^2+1=3+1=4\)

\(f\left(3\right)=3.\left(3\right)^2+1=27+1=28\)

1a.

\(y'=3x^2.f'\left(x^3\right)-2x.g'\left(x^2\right)\)

b.

\(y'=\dfrac{3f^2\left(x\right).f'\left(x\right)+3g^2\left(x\right).g'\left(x\right)}{2\sqrt{f^3\left(x\right)+g^3\left(x\right)}}\)

2.

\(f'\left(x\right)=\left(m-1\right)x^3+\left(m-2\right)x^2-2mx+3=0\)

Để ý rằng tổng hệ số của vế trái bằng 1 nên pt luôn có nghiệm \(x=1\), sử dụng lược đồ Hooc-ne ta phân tích được:

\(\Leftrightarrow\left(x-1\right)\left[\left(m-1\right)x^2+\left(2m-3\right)x-3\right]=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\\left(m-1\right)x^2+\left(2m-3\right)x-3=0\left(1\right)\end{matrix}\right.\)

Xét (1), với \(m=1\Rightarrow x=-3\)

- Với \(m\ne1\Rightarrow\Delta=\left(2m-3\right)^2+12\left(m-1\right)=4m^2-3\)

Nếu \(\left|m\right|< \dfrac{\sqrt{3}}{2}\Rightarrow\) (1) vô nghiệm \(\Rightarrow f'\left(x\right)=0\) có đúng 1 nghiệm

Nếu \(\left|m\right|>\dfrac{\sqrt{3}}{2}\Rightarrow\left(1\right)\) có 2 nghiệm \(\Rightarrow f'\left(x\right)=0\) có 3 nghiệm

2: ĐKXĐ: x<>1

\(f'\left(x\right)=\dfrac{\left(x^2-3x+3\right)'\left(x-1\right)-\left(x^2-3x+3\right)\left(x-1\right)'}{\left(x-1\right)^2}\)

\(=\dfrac{\left(2x-3\right)\left(x-1\right)-\left(x^2-3x+3\right)}{\left(x-1\right)^2}\)

\(=\dfrac{2x^2-5x+3-x^2+3x-3}{\left(x-1\right)^2}=\dfrac{x^2-2x}{\left(x-1\right)^2}\)

f'(x)=0

=>x^2-2x=0

=>x(x-2)=0

=>\(\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)

1:

\(f\left(x\right)=\dfrac{1}{3}x^3-2\sqrt{2}\cdot x^2+8x-1\)

=>\(f'\left(x\right)=\dfrac{1}{3}\cdot3x^2-2\sqrt{2}\cdot2x+8=x^2-4\sqrt{2}\cdot x+8=\left(x-2\sqrt{2}\right)^2\)

f'(x)=0

=>\(\left(x-2\sqrt{2}\right)^2=0\)

=>\(x-2\sqrt{2}=0\)

=>\(x=2\sqrt{2}\)